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Question
In linux, assume you are within a shell script, print out the number of command-line arguments fed to the script.
Answer

echo $#

^^ note that $ denotes variable expansion, and # is special variable to represent number of command-line arguments to current script


Question
In linux, assume you are within a shell script, print out the number of command-line arguments fed to the script.
Answer
?

Question
In linux, assume you are within a shell script, print out the number of command-line arguments fed to the script.
Answer

echo $#

^^ note that $ denotes variable expansion, and # is special variable to represent number of command-line arguments to current script

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20. Advanced Shell Scripting
2" ... When there are no positional parameters, "$@" and $@ expand to nothing (i.e., they are removed). [Hint: this is very useful for writing wrapper shell scripts that just add one argument.] <span>$# Expands to the number of positional parameters in decimal (i.e. the number of command-line arguments). $? Expands to the status of the most recently executed foreground pipeline. [I.e.,

Summary

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repetition number in this series0memorised on               scheduled repetition               
scheduled repetition interval               last repetition or drill

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